Answer

$y'=\dfrac{3x^2-14x}{x^3-7x^2-3}$

Work Step by Step

In order to derivate this function you have to apply the chain rule
Let's make a «u» substitution to make it easier
$f(u) = \ln(u)$
$u = x^3-7x^2-3$
Derivate the function:
$f'(u) = \dfrac{1}{u} \times u'$
Now let's find u'
$u' =3x^2-14x$
Then undo the substitution, simplify and get the answer::
$y'=\dfrac{3x^2-14x}{x^3-7x^2-3}$